Probability 3 Question 34
34. If $A$ and $B$ are two independent events, prove that $P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(C)$, where $C$ is an event defined that exactly one of $A$ and $B$ occurs. $\quad$ (2004, 2M)
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Solution:
- Here, $P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right)$
$$ \Rightarrow \quad{P(A)+P(B)-P(A \cap B)}{P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) } $$
[since $A, B$ are independent, so $A^{\prime}, B^{\prime}$ are independent]
$\therefore P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq{P(A)+P(B)} \cdot{P\left(A^{\prime}\right) \cdot P(B) }$
$$ =P(A) \cdot P\left(A^{\prime}\right) \cdot P(B)+P(B) \cdot P\left(A^{\prime}\right) \cdot P(B) $$
$$ \leq P(A) \cdot P(B)+P(B) \cdot P\left(A^{\prime}\right) $$
$$ \left[\because P\left(A^{\prime}\right) \leq 1 \text { and } P\left(B^{\prime}\right) \leq 1\right] $$
$\Rightarrow \quad P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(A) \cdot P\left(B^{\prime}\right)+P(B) \cdot P\left(A^{\prime}\right)$
$\Rightarrow P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(C)$
$$ \left[\because P(C)=P(A) \cdot P\left(B^{\prime}\right)+P(B) \cdot P\left(A^{\prime}\right)\right] $$
